Find the range of the function f(x) = (x)/(1 + x²) — JEE Mathematics

Let $y = \frac{x}{1 + x^2} \implies y + yx^2 = x \implies yx^2 - x + y = 0$.
For $x$ to be a real number, the discriminant of this quadratic equation in $x$ must be non-negative ($D \ge 0$).

$$(-1)^2 - 4(y)(y) \ge 0 \implies 1 - 4y^2 \ge 0 \implies 4y^2 \le 1 \implies y^2 \le \frac{1}{4}$$

Taking the square root:

$$-\frac{1}{2} \le y \le \frac{1}{2}$$

Thus, the range of the function is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.

Answer: $\left[-\frac{1}{2}, \frac{1}{2}\right]$